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For matrices of positive determinant, use the determinant-normalized slash operator for modular formsIt satisfies the composition law $(f{|}_k[\alpha ]){|}_k[\beta ]=f{|}_k[\alpha \beta ]$. IfthenThus the Fricke involution normalizes the group, and the composition law proves that $f{|}_k[{\alpha }_N]$ has the required transformation law.

The matrix ${\alpha }_N$ permutes the rational cusps. Applying it to a local Fourier expansion merely transports that expansion to the image cusp, with a nonzero change of local parameter. It therefore preserves holomorphy and vanishing at every cusp. Hence

LetForthe matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 3b\\ c& d\end{array}\right)$ lies in $S{L}_2(\mathbb{Z})$ and satisfies $3\gamma \tau ={\gamma }^{\prime }(3\tau )$. The weight-twelve transformation law for the modular discriminant gives the same factor $(3c\tau +d{)}^{12}$ in numerator and denominator. Hence $f(\gamma \tau )=f(\tau )$, so $f$ is a weight-zero modular function of level ${\Gamma }_0(3)$.

The discriminant has no zero in the upper half-plane, so $f$ has neither zeros nor poles there. At infinity,and therefore $f=q^{-2}+O(q^{-1})$: it has a pole of order two. The transformation $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$ givesThe Fricke involution exchanges infinity and zero, so $f$ has a zero of order two at the cusp zero.

Thus the morphism ${\varphi }_f:X_0(3)\to \hat{\mathbb{C}}$ has degree two, equal to its total pole order. An isomorphism of compact Riemann surfaces has degree one. Although $X_0(3)$ has genus zero, this particular morphism is therefore not an isomorphism.